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Q.Why a signal transmitted from a TV tower cannot be received beyond a certain distance? Write the expression for the optimum separation between the receiving and the transmitting antenna.

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★est
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The Earth's curvature blocks line-of-sight propagation, limiting the range of TV signals. The optimum separation between transmitting and receiving antennas is d=2RhT+2RhRd = \sqrt{2Rh_T} + \sqrt{2Rh_R}, where RR is the Earth's radius and hTh_T, hRh_R are the antenna heights.

The Concept: Why TV Signals Have a Range Limit

TV signals in the VHF and UHF bands (used for terrestrial broadcasting) travel in straight lines — they are line-of-sight (LOS) waves. Unlike lower-frequency radio waves that can bend around obstacles or follow the Earth's curvature (ground waves), these higher-frequency signals pass through the ionosphere without reflecting back and are not significantly diffracted by the Earth's bulge.

The Earth is roughly spherical. So even with a powerful transmitter, the signal cannot reach a receiver that lies below the horizon as seen from the transmitting antenna. The curvature of the Earth creates a radio horizon — beyond it, the signal is physically blocked.

This is why you cannot receive a TV tower's signal beyond a certain distance, no matter how strong the transmitter. The limiting factor is geometry, not power.

Watch out

A common mistake is to think the range is limited by signal attenuation or power loss. While those affect quality, the fundamental limit for VHF/UHF TV is the Earth's curvature blocking the line of sight.

Deriving the Optimum Separation

We want the maximum distance between a transmitting antenna (height hTh_T) and a receiving antenna (height hRh_R) such that they still have a direct line of sight, just grazing the Earth's surface.

  1. Set up the geometry.

    Consider the Earth as a perfect sphere of radius R≈6400 kmR \approx 6400\ \text{km}. Draw the transmitting antenna of height hTh_T at point TT, and the receiving antenna of height hRh_R at point RR. The line of sight from TT to RR just touches the Earth's surface at a point PP between them. The centres of the Earth, TT, PP, and RR all lie in the same vertical plane.

  2. Find the distance from the transmitter to the horizon.

    From the transmitter at height hTh_T, draw a tangent to the Earth's surface. This tangent touches the Earth at the horizon point. The distance dTd_T from the transmitter to this horizon point is the length of the tangent segment.

    Using the right-angled triangle formed by the Earth's centre OO, the transmitter TT, and the horizon point HTH_T:

    • OT=R+hTOT = R + h_T
    • OHT=ROH_T = R
    • THT=dTTH_T = d_T By Pythagoras:

(R+hT)2=R2+dT2(R + h_T)^2 = R^2 + d_T^2

R2+2RhT+hT2=R2+dT2R^2 + 2Rh_T + h_T^2 = R^2 + d_T^2

dT2=2RhT+hT2d_T^2 = 2Rh_T + h_T^2

Since hT≪Rh_T \ll R (antenna heights are tens to hundreds of metres, while RR is 6400 km), we neglect hT2h_T^2:

dT≈2RhTd_T \approx \sqrt{2Rh_T}

  1. Similarly, find the distance from the receiver to its horizon. …

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